Question #a621d

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Question #a621d

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Answer 1 · 2017-12-03T21:04:05

I am assuming missing parentheses to get $f (x) = a x e^{b x^{2}}$ $f(x) = axe^(bx^2)$ . Which makes $b = - \frac{1}{18}$ $b=-1/18$ and $a = \frac{7}{3} \sqrt{e}$ $a = 7/3sqrte$

Explanation:

$f'(x) = ae^(bx^2)(1+2bx^2) = 0$ at $x^2 = -1/(2b)$

So we need $x = sqrt(-1/(2b))=3$ and so,

$b = -1/18$

Now we have $f(x) = axe^(-x^2/18)$ .

We also want $f(x) = 7$ , so solve

$a(3)(e^(-1/2)) = 7$ for $a = 7/3e^(1/2)$

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Question #a621d

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